You are assigned the design of a cylindrical, pressurized water tank for a futur

ID: 1508594 • Letter: Y

Question

You are assigned the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 meters per second per second. The pressure at the surface of the water will be 150 kPa , and the depth of the water will be 14.1 m . The pressure of the air in the building outside the tank will be 87.0 kPa .

Part A

Find the net downward force on the tank's flat bottom, of area 1.65 m2 , exerted by the water and air inside the tank and the air outside the tank.

Express your answer numerically in Newtons, to three significant figures.

Part A

Find the net downward force on the tank's flat bottom, of area 1.65 m2 , exerted by the water and air inside the tank and the air outside the tank.

Express your answer numerically in Newtons, to three significant figures.

Explanation / Answer

Pressure at the surface = 150 kPa

height of water = 14.1 m

acceleration = 3.71 m/s2

density of water = 1000 kg/m3

So pressure at bottom of tank = pressure at top surface of water + pressure difference due to water column

= 150000 + 14.1 * 3.71 * 1000 = 202311 Pa

Area = 1.65 m2

So downward force at the bottom due to this = pressure at bottom * area = 202311 * 1.65 = 333813.15 N

Pressure applied by air column upwards on the bottom surface = 87 kPa

So upward force at the bottom due to this air column = pressure at bottom due air column * area = 87000 * 1.65 = 143550 N

Therefore the net downward force on tank's flat bottom = 333813.15 - 143550 = 190263.15 N = 1.90*105 N (in 3 significant figure)

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You are assigned the design of a cylindrical, pressurized water tank for a futur

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You are assigned the design of a cylindrical, pressurized water tank for a futur

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